Enzyme Kinetics and Inhibition

Question 1

Describe enzyme kinetics.

Answer 1

• Enzyme and substrate must combine to form ES complex, then enzyme must be recycled after the reaction is finished,
• Reaction at equilibrium with formation of ES complex, with rates k 1 and k -1
• Rate limiting step is production of product, driven by k 2 rate.
• As reaction is thermodynamically stable there is very little conversion of product back to substrate, so k -2 ignored

Question 2

What is the relationship between the concentration if a substrate and enzymatic rate?

Answer 2

• First Order: Rate dependent on [S]
• Zero Order: No relationship between V and [S]
• Second Order: Relationship between V and [S] not proportional to [S], but rather multiple substrates

Question 3

To study enzymes, first order kinetics must be present:

Answer 3

• Velocity increases as [S] increases (first order kinetics) up to a point where the enzyme is “saturated” with substrate (Vmax)
• At Vmax, rate of the reaction unaffected by increases [S] - all enzyme active sites in use. (zero order kinetics)

Question 4

Describe enzyme conc. and reaction rate.

Answer 4

• Rate of reaction increases as enzyme concentration increases (constant [S])
• At higher enzyme concentrations, greater availability to catalyse reaction
• Linear relationship between reaction rate and [E] (at constant [S])

Question 5

What does the machetes-menten equation assume?

Answer 5

• Equilibrium - the association and dissociation of the substrate and enzyme
  is assumed to be a rapid equilibrium.
  E+S ⟶ ES is fast
  ES ⟶ E+ P is rate limiting
• Steady state - ES immediately comes to steady state and is a constant.
  i.e. ES is formed as fast as enzyme releases the product.
• At early time points, at initial velocity (V0), [P] ≈ 0
• Enzyme exists in only two forms: free (E) and substrate-bound (ES)

Question 6

What is the Michaelis-menten equation?

Answer 6

V0 = Vmax[S]/Km+[S]

Question 7

What is the Michaelis Constant (Km)

Answer 7

Km = k-1 + k2/k1

Question 8

Describe enzymes characterised by Km

Answer 8

• [S] at which the rate of reaction is half its maximum (1/2 Vmax)
• Represents dissociation constant (substrate affinity) of ES.
• Low values indicate ES complex held together tightly and rarely
  dissociates without S first reacting to form P.

Question 9

What is the physiological relevance of Km?

Answer 9

Utilisation of glucose in the liver
* Glucose converted by two different kinases to form glucose-6-phosphate.
glucose + ATP > glucose 6-phosphate + ADP
KM for glucose: Hexokinase = 0.05 mM, Glucokinase = 0.5 mM.
* Low blood sugar (fasted state), hexokinase phosphorylates glucose;
* When blood glucose rises (feeding), the high KM enzyme also functions.

Question 10

Describe enzyme catalytic constant kcat

Answer 10

• At Vmax with high [S], rate determined by [E].
• Rate constant under these conditions is catalytic constant, kcat = k2
• kcat = turnover number
  = max number of S converted to P per second by each active site.
• Measures how fast a given enzyme can catalyze a specific reaction
  (units = s -1)
• Larger k cat = more rapid catalytic events at the enzyme’s active site

Question 11

What are the real world limitation of Michaelis-menten kinetics?

Answer 11

Michaelis-Menten kinetics relies on law of mass action: assumes free diffusion and thermodynamically-driven random collision.
Many biochemical or cellular processes deviate from these conditions:
* Cytoplasm behaves more like a gel than a freely flowable aqueous
solution, severely limiting molecular movements (diffusion or collision).
* Heterogeneous enzymatic reactions - molecular mobility of E or S can be
restricted, due to immobilisation or phase-separation of reactants.
e.g. membrane enzymes.
* Homogenous enzymatic reactions - mobility of E or S may be limited
e.g. DNA polymerase where E moves along a chained substrate,
rather than having a three-dimensional freedom.

Question 12

Describe random substrate binding (Ternary Complex).

Answer 12

• Assumes independent binding of substrates and products
• Two independent binding sites; Substrate binding independent of other substrate

Question 13

Describe the ordered substrate binding (Ternary complex)

Answer 13

• One substrate must bind before second substrate can bind effectively

Question 14

Describe the ping-pong mechanism.

Answer 14

• Enzyme binds substrate A and then releases P
• Intermediate form of enzyme (E*) often carries A fragment and then binds B
• Product Q released and Enzyme returns to original state (E)

Question 15

Describe competitive inhibitor.

Answer 15

• Reversible and has a structure similar to S
• Competes with S for active site
• Effect reversed by increasing [S]

Question 16

Describe noncompetitive inhibitors.

Answer 16

• Structure different to S; Binds to allosteric site
• Distorts the shape of E and active site, prevents S binding
• Not reversed by increasing [S]

Question 17

Describe uncompetitive inhibitors.

Answer 17

• Inhibitor binds only to ES complex, not free E.
• Reduction in effective [ES] increases E apparent affinity for S
• KM lowered, decreased Vmax
• Takes longer for the S or P to leave active site.
• Work best when [S] high.

Question 18

Describe how aspirin is an irreversible enzyme inhibition.

Answer 18

• Bind covalently to enzyme and permanently inhibit it, cannot be reversed
• Often contain reactive functional electrophilic groups, covalently bind to
  nucleophilic residues, such as Ser, Cys and Tyr

Question 19

Describe MONOAMINE OXIDASE inhibitors.

Answer 19

• Modified substrates
• Initially processed by E as if it were the normal S;
• Reaction intermediate covalently and irreversibly binds E causing inhibition
  MAO catalyses serotonin breakdown leads to high MAO activity = depression

Question 20

Describe penicillin in terms of irreversible enzyme inhibition

Answer 20

• Transition-state analogs bind more tightly to enzyme than substrate or product:
• Penicillin inhibits glycopeptidyl transferase,
  enzyme that synthesizes cross-links in bacterial cell wall.
• Kills growing cells by inactivating enzyme

Question 21

What is the reality for uncompetitive inhibition.

Answer 21

Inhibitor binding should only occur if the active site is occupied by substrate. But in most cases, the inhibitor will have some affinity for the unoccupied enzyme as well

Question 22

What is the reality of non-competitive inhibition

Answer 22

The inhibitor affinity should be unchanged regardless of whether substrate is bound or not. The affinity for the inhibitor usually changes when substrate is bound in
reality